Originally Posted by

**Jools** Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'

= -500(1 + e^-t)^-2(e^-t)(ln e)

From where did you bring that $\displaystyle \ln e=1$ ? It can only confuse you.

= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)

Here's the mistake: you actually have $\displaystyle P'(3)=-500(1+e^{-3})(-2e^{-3})=1,000(1+e^{-3})e^{-3}$ , which is positive indeed.

Tonio

In fact now I realized that that -2 is a power (!!), but still: you wrote $\displaystyle (1+e^{-t})'=e^{-t}$ , which is lacking a minus sign there!

Tonio

= -500(.9074)(.04979)

= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.